TS EAMCET 5 May 2018 Shift 1 Solved Paper

The kinetic energy of an emitted electron given is 1.986×10−19J
The frequency of radiation is 1×1015s−1
The threshold frequency (vo) can be calculated as follows:
hv−hvo=KE
h(v−vo)=KE
Substitute the values in the above formula.
6.62×10−34J.s(1×1015s−1−vo) =1.986×10−19J
1×1015s−1−v0=

1.986×10−19J

6.62×10−34J.s

vo=1×1015s−1−0.3×1015s−1
vo=7.0×1014s−1